A roller coaster car has a mass of 400kg and a speed of 15m/s What will be the P.E of this roller coaster car at its highest poi
1 answer:
Answer:
3000J
Explanation:
Given parameters:
mass = 400kg
speed = 15m/s
Unknown:
P.E at the highest point of the roller coaster = ?
Solution:
Due to the law of conservation of energy, the potential energy at the highest point can be solved using the formula of the kinetic energy
Potential energy = mv²
m is the mass
v is the velocity
Potential energy = x 400 x 15 = 3000J
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Answer:
Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g
Explanation:
The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.
And sum of upward forces = sum of downward forces.
N = mg cos θ
m = 100 g = 0.10 kg
g = acceleration due to gravity = 9.8 m/s²
θ = 15°
N = (0.1×9.8×cos 15°) = 0.946582 N
The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).
For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.
Frictional force = Tension + F
Frictional force = μN
where μ = coefficient of kinetic friction = 0.60
N = normal reaction = 0.9466 N
Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N
The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ
F = (0.10 × 9.8 × sin 15°) = 0.253624 N
Tension in the rope = T = ?
Fr = F + T
T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N
But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.
2T = mg
2 × 0.3144 = 9.8m
m = 0.06416 kg = 64.16 g.
Mass of the container = 30 g
So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g
Hope this Helps!!!
Answer:
The instantaneous speed of the object after the first five seconds is 12.5 m/s.
(C) is correct option.
Explanation:
Given that,
An object starts at rest. Its acceleration over 30 seconds.
We need to calculate the instantaneous speed of the object after the first five seconds
We know that,
Area under the acceleration -time graph gives speed.
According to figure,
Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.